1. We need the set $A=\{(x,y)\in \mathbb R^2: |x+y|\ge 3\}$ and $B=\{(x,y)\in \mathbb R^2: |x|+|y|\le 3\}.$

Also, $C=\{(x,y)\in A\cap B: x=0 \text{ or } y=0\}.$ We must find $\sum_{(x,y)\in C}|x+y|.$

Since the notation uses a sum over elements of $C$, the set $C$ must be finite. This happens because imposing $x=0$ or $y=0$ together with the two conditions gives only a few points.

2. Case 1: $x=0$

Then the conditions become:

• From $A$: $|x+y|=|y|\ge 3$
• From $B$: $|x|+|y|=|y|\le 3$

Combining, $|y|=3$ So, $y=\pm 3$ Hence the points are $(0,3),\ (0,-3).$

3. Case 2: $y=0$

Then the conditions become:

• From $A$: $|x+y|=|x|\ge 3$
• From $B$: $|x|+|y|=|x|\le 3$

Combining, $|x|=3$ So, $x=\pm 3$ Hence the points are $(3,0),\ (-3,0).$

4. Therefore, $C=\{(0,3),(0,-3),(3,0),(-3,0)\}.$

Now compute $|x+y|$ for each point:

• For $(0,3)$: $|x+y|=|3|=3$
• For $(0,-3)$: $|x+y|=|-3|=3$
• For $(3,0)$: $|x+y|=|3|=3$
• For $(-3,0)$: $|x+y|=|-3|=3$

So, $\sum_{(x,y)\in C}|x+y|=3+3+3+3=12.$

5. Comparing with the options:

• A: $18$
• B: $24$
• C: $15$
• D: $12$

Hence the correct option is $\boxed{\text{D}}.$